De moirve's
(√3-i ÷ √3+i)^6 = 1

(√3 - i ) / (√3 + i ) × (√3 - i ) / (√3 - i ) = (√3 - i )² / ((√3)² - i ²)

… = ((√3)² - 2√3 i + i ²) / (3 - i ²)

… = (3 - 2√3 i - 1) / (3 - (-1))

… = (2 - 2√3 i ) / 4

… = 1/2 - √3/2 i

… = √((1/2)² + (-√3/2)²) exp(i arctan((-√3/2)/(1/2))

… = exp(i arctan(-√3))

… = exp(-i arctan(√3))

… = exp(-iπ/3)

By DeMoivre's theorem,

[(√3 - i ) / (√3 + i )]⁶ = exp(-6iπ/3) = exp(-2iπ) = 1